Dolbier HW Solutions 410

Dolbier HW Solutions 410 - constant K a , which is 1.4 3 10...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
__ __ __ 504 CARBOXYLIC ACIDS 19.4 ( b ) Propanoic acid is similar to acetic acid in its acidity. A hydroxyl group at C-2 is electron- withdrawing and stabilizes the carboxylate ion of lactic acid by a combination of inductive and f eld effects. Lactic acid is more acidic than propanoic acid. The measured ionization constants are ( c ) A carbonyl group is more strongly electron-withdrawing than a carbon carbon double bond. Pyruvic acid is a stronger acid than acrylic acid. ( d ) Viewing the two compounds as substituted derivatives of acetic acid, RCH 2 CO 2 H, we judge to be strongly electron-withdrawing and acid-strengthening, whereas an ethyl group has only a small effect. 19.5 The compound can only be a carboxylic acid; no other class containing only carbon, hydrogen, and oxygen is more acidic. A reasonable choice is HC > CCO 2 H; C-2 is sp -hybridized and therefore rather electron-withdrawing and acid-strengthening. This is borne out by its measured ionization
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: constant K a , which is 1.4 3 10 2 2 (p K a 1.8). 19.6 For carbonic acid, the true K 1 is given by True K 1 5 } [H 1 [H ][ 2 H C C O O 3 ] 3 2 ] } The observed K is given by the expression 4.3 3 10 2 7 5 [H 1 ][HCO 3 2 ] } [CO 2 ] Butanoic acid K a 1.5 3 10 2 5 (p K a 4.7) CH 3 CH 2 CH 2 CO 2 H Methanesulfonylacetic acid K a 4.3 3 10 2 3 (p K a 2.4) CH 3 SCH 2 CO 2 H O O CH 3 S O O Pyruvic acid K a 5.1 3 10 2 4 (p K a 3.3) CH 3 CCO 2 H O Acrylic acid K a 5.5 3 10 2 5 (p K a 4.3) H 2 C CHCO 2 H Propanoic acid K a 1.3 3 10 2 5 (p K a 4.9) CH 3 CH 2 CO 2 H Lactic acid K a 1.4 3 10 2 4 (p K a 3.8) CH 3 CHCO 2 H OH Hydroxyl group stabilizes negative charge by attracting electrons. CH 3 CH C O OH O 2...
View Full Document

Ask a homework question - tutors are online